1) \(2x-\frac{4}{3}-\frac{4}{15}-\frac{4}{35}-\frac{4}{63}-\frac{4}{99}=\frac{15}{17}\)
2)\(\frac{10}{1\cdot2\cdot3}+\frac{10}{2\cdot3\cdot4}+\frac{10}{3\cdot4\cdot5}+.....+\frac{10}{100\cdot101\cdot102}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
9 tháng 5 2019
\(\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}\right)\cdot10-x=10\)10
\(\left(1-\frac{1}{10}\right)\cdot10-10=x\)
\(x=10\cdot\left(1-\frac{1}{10}-1\right)\)
\(x=10\cdot-\frac{1}{10}=-1\)
9 tháng 5 2019
\(\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\right).10-x=10\)
\(\left[1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right].10-x=10\)
\(\left[1+\left(\frac{1}{2}-\frac{1}{2}\right)+\left(\frac{1}{3}-\frac{1}{3}\right)+...+\left(\frac{1}{9}-\frac{1}{9}\right)-\frac{1}{10}\right].10-x=10\)
\(\left[1-\frac{1}{10}\right].10-x=10\)
\(\frac{9}{10}.10-x=10\)
\(9-x=10\)
\(x=9-10\)
\(x=-1\)
~ Hok tốt ~
22 tháng 6 2017
a)\(\frac{13}{15}+\frac{13}{35}+\frac{13}{63}+\frac{13}{99}\)
\(=\frac{13}{3.5}+\frac{13}{5.7}+\frac{13}{7.9}+\frac{13}{9.11}\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{5}+.....+\frac{1}{9}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\left(\frac{1}{3}-\frac{1}{11}\right)\)
\(=\frac{13}{2}\cdot\frac{8}{33}\)
\(=\frac{52}{33}\)
22 tháng 6 2017
a) Đặt A= 13/15 + 13/35 + 13/63 + 13/99
A = 13/2 ( 2/15 + 2/35 + 2/63 + 2/99)
A= 13/2 ( 2/ 3.5 + 2/5.7 + 2/7.9 + 2/9.11)
A= 13/2 ( 1/3 - 1/5 + 1/5 - 1/7 + 1/7 - 1/9 + 1/9 - 1/11)
A= 13/2 ( 1/3 - 1/11)
A= 13/2 . 8/33
A= 52/33
2 tháng 8 2015
A=\(\frac{1.2.3.4...8.9}{2.3.4.5...9.10}\)
A=\(\frac{1}{10}\)
mình làm đc 1 câu thôi. Bạn thông cảm nhé
23 tháng 9 2018
\(C=\frac{1.2-1}{2!}+\frac{2.3-1}{3!}+....+\frac{99.100-1}{100!}\)
\(\Rightarrow C=\frac{1.2}{2!}-\frac{1}{2!}+\frac{2.3}{3!}-\frac{1}{3!}+...+\frac{99.100}{100!}-\frac{1}{100!}\)
\(\Rightarrow C=\left(\frac{1.2}{2!}+\frac{2.3}{3!}+...+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(\Rightarrow C=\left(2+\frac{3.4}{4!}+\frac{4.5}{5!}+....+\frac{99.100}{100!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{10!}\right)\)
\(\Rightarrow C=\left(2+\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{98!}\right)-\left(\frac{1}{2!}+\frac{1}{3!}+...+\frac{1}{100!}\right)\)
\(\Rightarrow C=2-\frac{1}{99!}-\frac{1}{100!}< 2\Rightarrow C< 2\)
\(b,C=\frac{3}{1^2.2^2}+\frac{5}{2^2.3^2}+....+\frac{19}{9^2.10^2}\)
\(\Rightarrow C=\frac{3}{\left(1.2\right)\left(1.2\right)}+\frac{5}{\left(2.3\right)\left(2.3\right)}+...+\frac{19}{\left(9.10\right)\left(9.10\right)}\)
\(\Rightarrow C=\frac{3}{1.2}.\frac{1}{1.2}+\frac{5}{2.3}.\frac{1}{2.3}+....+\frac{19}{9.10}.\frac{1}{9.10}\)
\(\Rightarrow C=\left(1+\frac{1}{2}\right)\left(1-\frac{1}{2}\right)+\left(\frac{1}{2}+\frac{1}{3}\right)\left(\frac{1}{2}-\frac{1}{3}\right)+....+\left(\frac{1}{9}+\frac{1}{10}\right)\left(\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+....+\frac{1}{81}-\frac{1}{90}\)
\(\Rightarrow C=1-\frac{1}{90}< 1\Rightarrow C< 1\)
19 tháng 7 2017
Ta có : \(\frac{1}{10.9}-\frac{1}{9.8}-.....-\frac{1}{2.1}\)
\(=\frac{1}{90}-\left(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+.....+\frac{1}{9.8}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....+\frac{1}{8}-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\left(1-\frac{1}{9}\right)\)
\(=\frac{1}{90}-\frac{8}{9}=\frac{-79}{90}\)
NV
Nguyễn Việt Lâm
Giáo viên
1 tháng 3 2020
\(a=lim\frac{\left(\frac{2}{3}\right)^n+1}{3\left(\frac{1}{3}\right)^n-12}=-\frac{1}{12}\)
\(b=lim\frac{4\left(\frac{4}{10}\right)^n+1}{\left(\frac{3}{10}\right)^n-40}=-\frac{1}{40}\)
\(c=lim\frac{1-\left(\frac{2}{12}\right)^n}{1+45\left(\frac{3}{12}\right)^n}=\frac{1}{1}=1\)
\(d=\frac{\left(-\frac{2}{3}\right)^n+1}{-2\left(-\frac{2}{3}\right)^n-12+2\left(\frac{1}{3}\right)^n}=-\frac{1}{12}\)
\(e=\frac{1-11\left(\frac{1}{3}\right)^n}{\left(\frac{1}{3}\right)^n+14\left(\frac{2}{3}\right)^n}=\frac{1}{0}=+\infty\)
\(f=\frac{\left(\frac{2}{5}\right)^n-3+\left(\frac{1}{5}\right)^n}{3\left(\frac{2}{5}\right)^n+28\left(\frac{4}{5}\right)^n}=\frac{-3}{0}=-\infty\)
6 tháng 8 2018
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
2, \(\frac{10}{1.2.3}+\frac{10}{2.3.4}+\frac{10}{3.4.5}+....+\frac{10}{100.101.102}\)
\(=\frac{3-1}{1.2.3}+\frac{4-2}{2.3.4}+\frac{5-3}{3.4.5}+...+\frac{102-100}{100.101.102}\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{2.3}+\frac{1}{2.3}-\frac{1}{3.4}+\frac{1}{3.4}-\frac{1}{4.5}+...+\frac{1}{100.101}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\left(\frac{1}{1.2}-\frac{1}{101.102}\right)\)
\(=\frac{10}{2}.\frac{2575}{5151}\)
\(=2,499514657\)
= 2,499514657 bạn nhé